A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{6}\]
Correct Answer: C
Solution :
Let the roots of the equation be\[p,\,\,q\]. Let \[S={{p}^{2}}+{{q}^{2}}\] \[={{(p+q)}^{2}}-2pq\] ... (i) Given equation is \[{{x}^{2}}-(\sin \alpha -2)x-(1+\sin \alpha )=0\] \[\therefore \] \[p+q=(\sin \alpha -2),\,\,pq=-(1+\sin \alpha )\] From Eq. (i), \[S={{(\sin \alpha -2)}^{2}}+2(1+\sin \alpha )\] \[={{\sin }^{2}}\alpha -4\sin \alpha +4+2+2\sin \alpha \] \[={{\sin }^{2}}\alpha -2\sin \alpha +6\] \[\Rightarrow \] \[S={{(\sin \alpha -1)}^{2}}+5\] This is least when\[\sin \alpha -1=0\] \[\Rightarrow \] \[\alpha =\frac{\pi }{2}\]
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